A line with slope 6 bisects the area of a unit square with vertices (1,0), (0,0) , (1,1), and (0,1). What is the y-intercept of this line?
I tried putting one point where the line intersects the square as (y,1), and the other as (x,0), and the y intercept as (0,a). Then, I tried using the slope to figure out the variables. The only problem is that I don't have enough equations. Help?
I would say it hit the square at (a, 0) and at (b , 1)
Area of trapezoid on left = (a+b)/2(1) = (a+b)/2
Area of trapezoid on right = [(1-b)+(1-a)]/2
= (2 - a - b)/2
if those areas are equal then
a + b = 2 - a - b
or as we could see from the sketch
a + b = 1
Now the slope of 6
b = a + (1/6)
so
a + a + 1/6 = 1
2 a = 5/6
a = 5/12
then
b = 5/12 + 1/6 = 7/12
NOW you have a problem I am sure you can do
find c in y = 6 x + c
It goes through (5/12 , 0)
0 = 5/2 + c
so
c = -5/2
y = 6 x - 5/2
for y axis intercept, x = 0
y = -5/2
You know the line is y=6(x-a) so it intercepts the x-axis at (0,a), and the line y=1 at x=(1+6a)/6 = 1/6 + a
So, the left half of the figure is a trapezoid with height 1 and bases a and (1/6 + a)
So, if that has area 1/2,
1(a+a+1/6)/2 = 1/2
a = 5/12
so, y = 6(x-5/12)
y = 6x - 5/2
---------------------------------
Or, consider the point C=(1/2,1/2), the center of the square. A vertical line through C would cut the square in half.
Since we want a line of slope 6, so y increases 6 for every increase in 1 by x. If the line passes through C, then if it leans to the right 1/12 at y=1 and to the left 1/12 at y=0, it will still cut the square in half. So, we have the line through (5/12,0) and (7/12,1).
y = 6x - 5/2
THANKS A LOT!
To find the y-intercept of the line, we can first find the equation of the line using the given slope. Then, we can find the x-coordinate of the point where the line intersects one of the sides of the unit square. Finally, we can substitute this x-coordinate into the equation of the line to find the y-intercept.
Let's start by finding the equation of the line with a slope of 6. The slope-intercept form of the equation of a line is given by y = mx + b, where m is the slope and b is the y-intercept.
Since the slope is given as 6, we have y = 6x + b.
Now, let's find the x-coordinate of the point where the line intersects one of the sides of the unit square. Since the line bisects the area of the square, its midpoint must lie on the line. The coordinates of the midpoint can be found by taking the average of the coordinates of the opposite corners of the square.
The opposite corners of the square are (1,0) and (0,1). Their average gives us the coordinates of the midpoint:
Midpoint x-coordinate = (1 + 0) / 2 = 0.5
Next, substitute this x-coordinate (0.5) into the equation of the line to solve for the y-coordinate of the midpoint:
y = 6(0.5) + b
y = 3 + b
Since the midpoint lies on the line, we can substitute the coordinates of the midpoint (0.5, y) into the equation of the line:
y = 6(0.5) + b
y = 3 + b
Now, we have two equations with two variables (y and b):
(1) y = 3 + b
(2) y = 6(0.5) + b
Solving these equations simultaneously will give us the values of y and b. Subtracting equation (1) from equation (2), we get:
0 = 3 - 6(0.5)
0 = 3 - 3
0 = 0
The equation is satisfied, which means the system of equations is dependent, and there are infinitely many solutions. This implies that the y-intercept (b) can be any value.
Therefore, the y-intercept of the line is not uniquely determined and can take on any real number value.